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Vilka [71]
3 years ago
6

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

k of inertia m is pressed against the free end of the spring until the spring is compressed a distance d from its relaxed length. Call this position A. The block is then released and moves up the ramp until coming to rest at position B. The surface is rough from position A for a distance 2d up the ramp, and over this distance the coefficient of kinetic friction for the two surfaces is μ . Friction is negligible elsewhere.
What is the distance from A to B? Suppose the values of μ, k, d, θ, and m are such that the spring fully extends, but the block never goes higher than 2d .
Express your answer in terms of some or all of the variables k, μ, m, d, and θ.


More than anything I'm just confused by the question. If someone could at least draw the picture I might be able to figure it out. Thanks.

Physics
1 answer:
Nikitich [7]3 years ago
5 0

Answer:

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

Explanation:

From the law of conservation of energy

Energy lost  by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x

x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

The required distance from A to B is x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

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what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?
zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

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